void SPL(int n, double *x, double *y, int ni, double *xi, double *yi)； 是你所要。
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已知 n 個點 x,y; x 必須已按順序排好。要插值 ni 點，橫坐標(biāo) xi[], 輸出 yi[]。
程序里用double 型，保證計算精度。
SPL調(diào)用現(xiàn)成的程序。
現(xiàn)成的程序很多。端點處理方法不同，結(jié)果會有不同。想同matlab比較，你需 嘗試 調(diào)用 spline（）函數(shù) 時，令 end1 為 1, 設(shè) slope1 的值，令 end2 為 1 設(shè) slope2 的值。
#include stdio.h
#include math.h
int spline (int n, int end1, int end2,
double slope1, double slope2,
double x[], double y[],
double b[], double c[], double d[],
int *iflag)
{
int nm1, ib, i, ascend;
double t;
nm1 = n - 1;
*iflag = 0;
if (n 2)
{ /* no possible interpolation */
*iflag = 1;
goto LeaveSpline;
}
ascend = 1;
for (i = 1; i n; ++i) if (x[i] = x[i-1]) ascend = 0;
if (!ascend)
{
*iflag = 2;
goto LeaveSpline;
}
if (n = 3)
{
d[0] = x[1] - x[0];
c[1] = (y[1] - y[0]) / d[0];
for (i = 1; i nm1; ++i)
{
d[i] = x[i+1] - x[i];
b[i] = 2.0 * (d[i-1] + d[i]);
c[i+1] = (y[i+1] - y[i]) / d[i];
c[i] = c[i+1] - c[i];
}
/* ---- Default End conditions */
b[0] = -d[0];
b[nm1] = -d[n-2];
c[0] = 0.0;
c[nm1] = 0.0;
if (n != 3)
{
c[0] = c[2] / (x[3] - x[1]) - c[1] / (x[2] - x[0]);
c[nm1] = c[n-2] / (x[nm1] - x[n-3]) - c[n-3] / (x[n-2] - x[n-4]);
c[0] = c[0] * d[0] * d[0] / (x[3] - x[0]);
c[nm1] = -c[nm1] * d[n-2] * d[n-2] / (x[nm1] - x[n-4]);
}
/* Alternative end conditions -- known slopes */
if (end1 == 1)
{
b[0] = 2.0 * (x[1] - x[0]);
c[0] = (y[1] - y[0]) / (x[1] - x[0]) - slope1;
}
if (end2 == 1)
{
b[nm1] = 2.0 * (x[nm1] - x[n-2]);
c[nm1] = slope2 - (y[nm1] - y[n-2]) / (x[nm1] - x[n-2]);
}
/* Forward elimination */
for (i = 1; i n; ++i)
{
t = d[i-1] / b[i-1];
b[i] = b[i] - t * d[i-1];
c[i] = c[i] - t * c[i-1];
}
/* Back substitution */
c[nm1] = c[nm1] / b[nm1];
for (ib = 0; ib nm1; ++ib)
{
i = n - ib - 2;
c[i] = (c[i] - d[i] * c[i+1]) / b[i];
}
b[nm1] = (y[nm1] - y[n-2]) / d[n-2] + d[n-2] * (c[n-2] + 2.0 * c[nm1]);
for (i = 0; i nm1; ++i)
{
b[i] = (y[i+1] - y[i]) / d[i] - d[i] * (c[i+1] + 2.0 * c[i]);
d[i] = (c[i+1] - c[i]) / d[i];
c[i] = 3.0 * c[i];
}
c[nm1] = 3.0 * c[nm1];
d[nm1] = d[n-2];
}
else
{
b[0] = (y[1] - y[0]) / (x[1] - x[0]);
c[0] = 0.0;
d[0] = 0.0;
b[1] = b[0];
c[1] = 0.0;
d[1] = 0.0;
}
LeaveSpline:
return 0;
}
double seval (int n, double u,
double x[], double y[],
double b[], double c[], double d[],
int *last)
{
int i, j, k;
double w;
i = *last;
if (i = n-1) i = 0;
if (i 0) i = 0;
if ((x[i] u) || (x[i+1] u))
{
i = 0;
j = n;
do
{
k = (i + j) / 2;
if (u x[k]) j = k;
if (u = x[k]) i = k;
}
while (j i+1);
}
*last = i;
w = u - x[i];
w = y[i] + w * (b[i] + w * (c[i] + w * d[i]));
return (w);
}
void SPL(int n, double *x, double *y, int ni, double *xi, double *yi)
{
double *b, *c, *d;
int iflag,last,i;
b = (double *) malloc(sizeof(double) * n);
c = (double *)malloc(sizeof(double) * n);
d = (double *)malloc(sizeof(double) * n);
if (!d) { printf("no enough memory for b,c,d\n");}
else {
spline (n,0,0,0,0,x,y,b,c,d,iflag);
if (iflag==0) printf("I got coef b,c,d now\n"); else printf("x not in order or other error\n");
for (i=0;ini;i++) yi[i] = seval(ni,xi[i],x,y,b,c,d,last);
free(b);free(c);free(d);
};
}
main(){
double x[6]={0.,1.,2.,3.,4.,5};
double y[6]={0.,0.5,2.0,1.6,0.5,0.0};
double u[8]={0.5,1,1.5,2,2.5,3,3.5,4};
double s[8];
int i;
SPL(6, x,y, 8, u, s);
for (i=0;i8;i++) printf("%lf %lf \n",u[i],s[i]);
return 0;
}
#include?stdio.h
double?Lerp(double?x0,double?y0,double?x1,double?y1,double?x)
{
double?dy?=?y1?-?y0;
if(dy?==?0){
printf("除0錯誤！\n");
return?0;
}
return?x?*?(x1?-?x0)?/?dy;
}
int?main()
{
double?x0,x1,y1,y0,x,y;
printf("Inptu?x0?y0?x1?y1?x:");
scanf("%lf?%lf?%lf?%lf?%lf",x0,y0,x1,y1,x);
y?=?Lerp(x0,y0,x1,y1,x);
printf("y?=?%lf\n",y);
return?0;
}
#includestdio.h
#includestdlib.h
#includeiostream.h
typedef struct data
{
float x;
float y;
}Data;//變量x和函數(shù)值y的結(jié)構(gòu)
Data d[20];//最多二十組數(shù)據(jù)
float f(int s,int t)//牛頓插值法，用以返回插商
{
if(t==s+1)
return (d[t].y-d[s].y)/(d[t].x-d[s].x);
else
return (f(s+1,t)-f(s,t-1))/(d[t].x-d[s].x);
}
float Newton(float x,int count)
{
int n;
while(1)
{
cout"請輸入n值(即n次插值):";//獲得插值次數(shù)
cinn;
if(n=count-1)// 插值次數(shù)不得大于count－1次
break;
else
system("cls");
}
//初始化t，y，yt。
float t=1.0;
float y=d[0].y;
float yt=0.0;
//計算y值
for(int j=1;j=n;j++)
{
t=(x-d[j-1].x)*t;
yt=f(0,j)*t;
//coutf(0,j)endl;
y=y+yt;
}
return y;
}
float lagrange(float x,int count)
{
float y=0.0;
for(int k=0;kcount;k++)//這兒默認(rèn)為count－1次插值
{
float p=1.0;//初始化p
for(int j=0;jcount;j++)
{//計算p的值
if(k==j)continue;//判斷是否為同一個數(shù)
p=p*(x-d[j].x)/(d[k].x-d[j].x);
}
y=y+p*d[k].y;//求和
}
return y;//返回y的值
}
void main()
{
float x,y;
int count;
while(1)
{
cout"請輸入x[i],y[i]的組數(shù)，不得超過20組:";//要求用戶輸入數(shù)據(jù)組數(shù)
cincount;
if(count=20)
break;//檢查輸入的是否合法
system("cls");
}
//獲得各組數(shù)據(jù)
for(int i=0;icount;i++)
{
cout"請輸入第"i+1"組x的值:";
cind[i].x;
cout"請輸入第"i+1"組y的值:";
cind[i].y;
system("cls");
}
cout"請輸入x的值:";//獲得變量x的值
cinx;
while(1)
{
int choice=3;
cout"請您選擇使用哪種插值法計算："endl;
cout" (0):退出"endl;
cout" (1):Lagrange"endl;
cout" (2):Newton"endl;
cout"輸入你的選擇：";
cinchoice;//取得用戶的選擇項
if(choice==2)
{
cout"你選擇了牛頓插值計算方法，其結(jié)果為：";
y=Newton(x,count);break;//調(diào)用相應(yīng)的處理函數(shù)
}
if(choice==1)
{
cout"你選擇了拉格朗日插值計算方法，其結(jié)果為：";
y=lagrange(x,count);break;//調(diào)用相應(yīng)的處理函數(shù)
}
if(choice==0)
break;
system("cls");
cout"輸入錯誤!!!!"endl;
}
coutx" , "yendl;//輸出最終結(jié)果
}
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